#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#include <array>
#include <unordered_set>
#include <unordered_map>
#define rep(x, s, t) for(ll x = (s); (x) <= (t); (x)++)
#define per(x, s, t) for(ll x = (s); (x) >= (t); (x)--)
#define reps(x, s) for(ll x = 0; (x) < (ll)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define outl(...) dump_func(__VA_ARGS__)
#define outf(x) cout << fixed << setprecision(16) << (x) << endl
#define pb push_back
#define fi first
#define se second
#define inf 2e18
#define eps 1e-9
const double PI = 3.1415926535897932384626433;

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

map<ll, ll> mp;
ll a[55], b[55], m[55];
bool prime[100005];

ll gcd(ll a, ll b){
	if(b == 0) return a;
	return gcd(b, a%b);
}

//ax+by = gcd(a, b)を満たす(x, y)を求めgcd(a, b)を返す
ll extgcd(ll a, ll b, ll &x, ll &y)
{
	if(b == 0){
		x = 1, y = 0;
		return a;
	}
	ll xx, yy;
	ll d = extgcd(b, a%b, xx, yy);
	x = yy, y = xx-(a/b)*yy;
	return d;
}

//a^{-1} (mod m)を求める。存在しない場合(gcd(a, m)!=1)は-1を返す
ll mod_inverse(ll a, ll m)
{
	ll x, y;
	if(extgcd(a, m, x, y) != 1) return -1;
	return (x%m + m) % m;
}

//ax = b (mod m)を満たすx(mod m/gcd(a, m))を求める。存在しない場合(b%gcd(a, m)!=0)は(0, -1)を返す
P congruence(ll a, ll b, ll m)
{
	ll d = gcd(a, m);
	if(b % d) return make_pair(0, -1);
	a /= d, b /= d, m /= d;
	return make_pair(b * mod_inverse(a, m) % m, m);
}

//連立合同方程式a_i*x = b_i (mod m_i)(i = 1, 2, ..., n)の解(x, M)を求める。存在しない場合(0, -1)を返す
P SimultaneousCongruence(ll a[], ll b[], ll m[], ll n)
{
	ll x = 0, M = 1;
	for(int i = 1; i <= n; i++){
		P res = congruence(a[i]*M, (b[i]-a[i]*x%m[i]+m[i])%m[i], m[i]);
		if(res.second == -1) return res;
		x += M*res.first, M *= res.second;
	}
	return make_pair(x, M);
}

const int FACT_MAX = 10000005;
ll q[FACT_MAX], e[FACT_MAX];

ll modpow(ll a, ll n, ll mod)
{
	if(n == 0) return 1;
	if(n % 2){
		return ((a%mod) * (modpow(a, n-1, mod)%mod)) % mod;
	}
	else{
		return modpow((a*a)%mod, n/2, mod) % mod;
	}
}

void make_fact(ll p, ll mod)
{
	ll qval = 1, eval = 0;
	q[0] = 1, e[0] = 0;
	
	for(int i = 1; i < FACT_MAX; i++){
		ll t = i;
		while(t % p == 0){
			eval++;
			t /= p;
		}
		qval *= t, qval %= mod;
		q[i] = qval, e[i] = eval;
	}
}

ll comb(ll n, ll k, ll p, ll ex, ll mod)
{
	if(n < 0 || k < 0 || n < k) return 0;
	ll eval = e[n] - e[k] - e[n-k];
	if(eval >= ex) return 0;
	
	ll ret = q[n] * mod_inverse(q[k]*q[n-k]%mod, mod) % mod;
	ret *= modpow(p, eval, mod), ret %= mod;
	return ret;
}

ll n, k;
ll calc(ll p, ll ex, ll mod)
{
	make_fact(p, mod);
	
	//mod = p^exのときの答えを求める処理を書く
	
	return comb(n, k, p, ex, mod);
}

//Mを法とする
int main(void)
{
	cin >> n >> k;
	ll M = 100000000;
	for(int i = 2; i < 100005; i++){
		if(prime[i]) continue;
		for(int j = 2*i; j < 100005; j+=i) prime[j] = true;
	}
	for(int i = 2; i < 100005; i++){
		if(prime[i]) continue;
		while(M%i == 0){
			mp[i]++;
			M /= i;
		}
	}
	if(M > 1) mp[M]++;
	
	ll id = 0;
	for(auto it = mp.begin(); it != mp.end(); it++){
		id++;
		ll mod = 1;
		for(int i = 0; i < it->second; i++) mod *= it->first;
		a[id] = 1, b[id] = calc(it->first, it->second, mod), m[id] = mod;
	}
	printf("%08d\n", (int)SimultaneousCongruence(a, b, m, id).first);
	
	return 0;
}